We Keep Coding

Java regex for google maps url?

I want to parse all google map links inside a String. The format is as follows :
1st example
https://www.google.com/maps/place/white+house/#38.8976763,-77.0387185,17z/data=!3m1!4b1!4m5!3m4!1s0x89b7b7bcdecbb1df:0x715969d86d0b76bf!8m2!3d38.8976763!4d-77.0365298
https://www.google.com/maps/place/white+house/#38.8976763,-77.0387185,17z
https://www.google.com/maps/place//#38.8976763,-77.0387185,17z
https://maps.google.com/maps/place//#38.8976763,-77.0387185,17z
https://www.google.com/maps/place/#38.8976763,-77.0387185,17z
https://google.com/maps/place/#38.8976763,-77.0387185,17z
http://google.com/maps/place/#38.8976763,-77.0387185,17z
https://www.google.com.tw/maps/place/#38.8976763,-77.0387185,17z
These are all valid google map URLs (linking to White House)
Here is what I tried
String gmapLinkRegex = "(http|https)://(www\\.)?google\\.com(\\.\\w*)?/maps/(place/.*)?#(.*z)[^ ]*";
Pattern patternGmapLink = Pattern.compile(gmapLinkRegex , Pattern.CASE_INSENSITIVE);
Matcher m = patternGmapLink.matcher(s);
while (m.find()) {
logger.info("group0 = {}" , m.group(0));
String place = m.group(4);
place = StringUtils.stripEnd(place , "/"); // remove tailing '/'
place = StringUtils.stripStart(place , "place/"); // remove header 'place/'
logger.info("place = '{}'" , place);
String latLngZ = m.group(5);
logger.info("latLngZ = '{}'" , latLngZ);
}
It works in simple situation , but still buggy ...
for example
It need post-process to grab optional place information
And it cannot extract one line with two urls such as :
s = "https://www.google.com/maps/place//#38.8976763,-77.0387185,17z " +
" and http://google.com/maps/place/#38.8976763,-77.0387185,17z";
It should be two urls , but the regex matches the whole line ...
The points :
The whole URL should be matched in group(0) (including the tailing data part in 1st example),
in the 1st example , if the zoom level : 17z is removed , it is still a valid gmap URL , but my regex cannot match it.
Easier to extract optional place info
Lat / Lng extraction is must , zoom level is optional.
Able to parse multiple urls in one line
Able to process maps.google.com(.xx)/maps , I tried (www|maps\.)? but seems still buggy
Any suggestion to improve this regex ? Thanks a lot !

The dot-asterisk
.*
will always allow anything to the end of the last url.
You need "tighter" regexes, which match a single URL but not several with anything in between.
The "[^ ]*" might include the next URL if it is separated by something other than " ", which includes line break, tab, shift-space...
I propose (sorry, not tested on java), to use "anything but #" and "digit, minus, comma or dot" and "optional special string followed by tailored charset, many times".
"(http|https)://(www\.)?google\.com(\.\w*)?/maps/(place/[^#]*)?#([0123456789\.,-]*z)(\/data=[\!:\.\-0123456789abcdefmsx]+)?"
I tested the one above on a perl-regex compatible engine (np++).
Please adapt yourself, if I guessed anything wrong. The explicit list of digits can probably be replaced by "\d", I tried to minimise assumptions on regex flavor.
In order to match "URL" or "URL and URL", please use a variable storing the regex, then do "(URL and )*URL", replacing "URL" with regex var. (Asuming this is possible in java.) If the question is how to then retrieve the multiple matches: That is java, I cannot help. Let me know and I delete this answer, not to provoke deserved downvotes ;-)
(Edited to catch the data part in, previously not seen, first example, first line; and the multi URLs in one line.)

I wrote this regex to validate google maps links:
"(http:|https:)?\\/\\/(www\\.)?(maps.)?google\\.[a-z.]+\\/maps/?([\\?]|place/*[^#]*)?/*#?(ll=)?(q=)?(([\\?=]?[a-zA-Z]*[+]?)*/?#{0,1})?([0-9]{1,3}\\.[0-9]+(,|&[a-zA-Z]+=)-?[0-9]{1,3}\\.[0-9]+(,?[0-9]+(z|m))?)?(\\/?data=[\\!:\\.\\-0123456789abcdefmsx]+)?"
I tested with the following list of google maps links:
String location1 = "http://www.google.com/maps/place/21.01196755,105.86306012";
String location2 = "https://www.google.com.tw/maps/place/#38.8976763,-77.0387185,17z";
String location3 = "http://www.google.com/maps/place/21.01196755,105.86306012";
String location4 = "https://www.google.com/maps/place/white+house/#38.8976763,-77.0387185,17z/data=!3m1!4b1!4m5!3m4!1s0x89b7b7bcdecbb1df:0x715969d86d0b76bf!8m2!3d38.8976763!4d-77.0365298";
String location5 = "https://www.google.com/maps/place/white+house/#38.8976763,-77.0387185,17z";
String location6 = "https://www.google.com/maps/place//#38.8976763,-77.0387185,17z";
String location7 = "https://maps.google.com/maps/place//#38.8976763,-77.0387185,17z";
String location8 = "https://www.google.com/maps/place/#38.8976763,-77.0387185,17z";
String location9 = "https://google.com/maps/place/#38.8976763,-77.0387185,17z";
String location10 = "http://google.com/maps/place/#38.8976763,-77.0387185,17z";
String location11 = "https://www.google.com/maps/place/#/data=!4m2!3m1!1s0x3135abf74b040853:0x6ff9dfeb960ec979";
String location12 = "https://maps.google.com/maps?q=New+York,+NY,+USA&hl=no&sll=19.808054,-63.720703&sspn=54.337928,93.076172&oq=n&hnear=New+York&t=m&z=10";
String location13 = "https://www.google.com/maps";
String location14 = "https://www.google.fr/maps";
String location15 = "https://google.fr/maps";
String location16 = "http://google.fr/maps";
String location17 = "https://www.google.de/maps";
String location18 = "https://www.google.com/maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4";
String location19 = "https://www.google.de/maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4";
String location20 = "https://www.google.com/maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4&layer=t&lci=com.panoramio.all,com.google.webcams,weather";
String location21 = "https://www.google.com/maps?ll=37.370157,0.615234&spn=45.047033,93.076172&t=m&z=4&layer=t";
String location22 = "https://www.google.com/maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4";
String location23 = "https://www.google.de/maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4";
String location24 = "https://www.google.com/maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4&layer=t&lci=com.panoramio.all,com.google.webcams,weather";
String location25 = "https://www.google.com/maps?ll=37.370157,0.615234&spn=45.047033,93.076172&t=m&z=4&layer=t";
String location26 = "http://www.google.com/maps/place/21.01196755,105.86306012";
String location27 = "http://google.com/maps/bylatlng?lat=21.01196022&lng=105.86298748";
String location28 = "https://www.google.com/maps/place/C%C3%B4ng+vi%C3%AAn+Th%E1%BB%91ng+Nh%E1%BA%A5t,+354A+%C4%90%C6%B0%E1%BB%9Dng+L%C3%AA+Du%E1%BA%A9n,+L%C3%AA+%C4%90%E1%BA%A1i+H%C3%A0nh,+%C4%90%E1%BB%91ng+%C4%90a,+H%C3%A0+N%E1%BB%99i+100000,+Vi%E1%BB%87t+Nam/#21.0121535,105.8443773,13z/data=!4m2!3m1!1s0x3135ab8ee6df247f:0xe6183d662696d2e9";

Error NumberFormatException on Java

I have the following exception trying to manipulate a value to be added;
thank you very much for your help
java.lang.NumberFormatException: Invalid double: "20,000"
java.lang.StringToReal.invalidReal(StringToReal.java:63)
java.lang.StringToReal.parseDouble(StringToReal.java:269)
java.lang.Double.parseDouble(Double.java:295)
java.lang.Double.valueOf(Double.java:332)

You can use a NumberFormat to parse your String1. Something like,
String str = "20,000";
NumberFormat nf = NumberFormat.getNumberInstance(new Locale("en_US"));
NumberFormat nfIT = NumberFormat.getNumberInstance(Locale.ITALIAN);
try {
System.out.println(nf.parse(str)); // <-- 20000
System.out.println(nfIT.parse(str)); // <-- 20
} catch (ParseException e) {
e.printStackTrace();
}
For more options see Customizing Formats (The Java Tutorials).
1Being sure to pass the appropriate Locale to match your expected output.

You don't use comma as a separator in numbers.
You have to use 20000 instead of 20,000.
EDIT:
as #MitchWeaver mentioned, you can also substitute comma to underescore, making it 20_000

Parsing a JSON Object in Java having special characters

I am stuck in a situation, where my JSONString (ruleFormJSONString) looks like :
{
"ruleDescription":"Test Rule2 Description",
"urlId":"1",
"listOfBusinessdays":["1","2","5","6","7"],
"status":"1",
"hierarchyId":"3",
"fromTime":"08:00",
"toTime":"18:00",
"dcnid":"1",
"eventId":"1",
"rowstate":"1",
"listOfLocations":["ASM","DEL"],
"ruleName":"Test Rule2",
"ruleId":"7","msgId":"1"
}
As you can see there are 2 attributes named fromTime and toTime which has a :
So while parsing this in Java, I used
JSONObject ruleFormJSON = JSONObject.fromString(ruleFormJSONString);
String fromTime = (String)ruleFormJSON.getString("fromTime");
String toTime = (String)ruleFormJSON.getString("toTime");
I am getting a NumberFormatException which is
java.lang.NumberFormatException: For input string: "18:00"
So please suggest me how, to get the value in the corresponding String variable.
Any help will be appreciated.

It seems there is an error on this line:
"listOfBusinessdays":"1","2","5","6","7"],
A closed bracket square but no open bracket before.
May be this hang up the parser.

Java use regex to extract file name

I need to get a file name from file's absolute path (I am aware of the file.getName() method, but I cannot use it here).
EDIT: I cannot use file.getName() because I don't need the file name only; I need the part of the file's path as well (but again, not the entire absoulte path). I need the part of file's path AFTER certain path provided.
Let's say the file is located in the folder:
C:\Users\someUser
On windows machine, if I make a pattern string as follows:
String patternStr = "C:\\Users\\someUser\\(.*+)";
I get an exception: java.util.regex.PatternSyntaxException: Illegal/unsupported escape sequence for backslash.
If I use Pattern.quote(File.pathSeparator):
String patternStr = "C:" + Pattern.quote(File.separator) + "Users" + Pattern.quote(File.separator) + "someUser" + Pattern.quote(File.separator) + "(.*+)";
the resulting pattern string is: C:\Q;\EUsers\Q;\EsomeUser\Q;\E(.*+) which of course has no match with the actual fileName "C:\Users\someUser\myFile.txt".
What am I missing here? What is the proper way to parse file name?

What is the proper way to parse file name?
The proper way to parse a file name is to use File(String). Using a regex for this is going to hard-wire platform dependencies into your code. That's a bad idea.
I know you said you can't use File.getName() ... but that is the proper solution. If you would care to say why you can't use File.getName() perhaps I could suggest an alternative solution.

If you indeed want to use a regular expressions, you should use
String patternStr = "C:\\\\Users\\\\someUser\\\\(.*+)";
^^ ^^ ^^
instead.
Why? Your string literal
"C:\\Users\\someUser\\(.*+)"
is compiled to
C:\Users\someUser\(.*+)
Since \ is used for escaping in regular expressions too, you'll have to escape them "twice".
Regarding your edit:
You probably want to have a look at URI.relativize(). Example:
File base = new File("C:/Users/someUser");
File file = new File("C:/Users/someUser/someDir/someFile.txt");
String relativePath = base.toURI().relativize(file.toURI()).getPath();
System.out.println(relativePath); // prints "someDir/someFile.txt"
(Note that / works as file-separator on Windows machines too.)
Btw, I don't know what you have as File.separator on your system, but if it's set to \, then
"C:" + Pattern.quote(File.separator) + "Users" + Pattern.quote(File.separator) +
"someUser" + Pattern.quote(File.separator) + "(.*+)";
should yield
C:\Q\\EUsers\Q\\EsomeUser\Q\\E(.*+)

String patternStr = "C:\\Users\\someUser\\(.*+)";
Backslashes (\) are escape characters in the Java Language. Your string contains the following after compilation:
C:\Users\someUser\(.*+)
This string is then parsed as a regex, which uses backslashes as an escape character as well. The regex parser tries to understand the escaped \U, \s and \(. One of them is incorrect regarding the regex syntax (hence your exception), and none of them are what you are trying to achieve.
Try
String patternStr = "C:\\\\Users\\\\someUser\\\\(.*+)";

If you want to solve it by pattern you need to escape your Pattern properly
String patternStr = "C:\\\\Users\\\\someUser\\\\(.*+)";

Try putting double-double-backslashes in your pattern. You need a second backslash to escape one in the patter, plus you'll need to double each one to escape them in the string. Hence you'll end up with something like:
String patternStr = "C:\\\\Users\\\\someUser\\\\(.*+)";

Move from end of string to first occurrence of file path separator* or begin.
File paths separator can be / or \.
public static final char ALTERNATIVE_DIRECTORY_SEPARATOR_CHAR = '/';
public static final char DIRECTORY_SEPARATOR_CHAR = '\\';
public static final char VOLUME_SEPARATOR_CHAR = ':';
public static String getFileName(String path) {
if(path == null || path.isEmpty()) {
return path;
}
int length = path.length();
int index = length;
while(--index >= 0) {
char c = path.charAt(index);
if(c == ALTERNATIVE_DIRECTORY_SEPARATOR_CHAR || c == DIRECTORY_SEPARATOR_CHAR || c == VOLUME_SEPARATOR_CHAR) {
return path.substring(index + 1, length);
}
}
return path;
}
Try to keep it simple ;-).

Try this :
String ResultString = null;
try {
Pattern regex = Pattern.compile("([^\\\\/:*?\"<>|\r\n]+$)");
Matcher regexMatcher = regex.matcher(subjectString);
if (regexMatcher.find()) {
ResultString = regexMatcher.group(1);
}
} catch (PatternSyntaxException ex) {
// Syntax error in the regular expression
}
Output :
myFile.txt
Also for input : C:/Users/someUser/myFile.txt
Output : myFile.txt

What am I missing here? What is the proper way to parse file name?
The proper way to parse a file name is to use the APIs that are already provided for the purpose. You've stated that you can't use File.getName(), without explanation. You are almost certainly mistaken about that.

I cannot use file.getName() because I don't need the file name only; I need the part of the file's path as well (but again, not the entire absoulte path).
OK. So what you want is something like this.
// Canonicalize paths to deal with ".", "..", symlinks,
// relative files and case sensitivity issues.
String directory = new File(someDirectory).canonicalPath();
String test = new File(somePathname).canonicalPath();
if (!directory.endsWith(File.separator)) {
directory += File.separator;
}
if (test.startsWith(directory)) {
String pathInDirectory = test.substring(directory.length()):
...
}
Advantages:
No regexes needed.
Doesn't break if the path separator is something other than \.
Doesn't break if there are symbolic links on the path.
Doesn't break due to case sensitivity issues.

Suppose the file name has special characters, specially when supporting MAC where special characters are allowing in filenames, server side Path.GetFileName(fileName) fails and throws error because of illegal characters in path. The following code using regex come for the rescue.
The following regex take care of 2 things
In IE, when file is uploaded, the file path contains folders aswell (i.e. c:\samplefolder\subfolder\sample.xls). Expression below will replace all folders with empty string and retain the file name
When used in Mac, filename is the only thing supplied as its safari browser and allows special chars in file name
var regExpDir = #"(^[\w]:\\)([\w].+\w\\)";
var fileName = Regex.Replace(fileName, regExpDir, string.Empty);

String format - java

What is the best way to replace 'gift' and 'price' from below string using java-
We’ve added {gift} coupon worth {price}. Enjoy!
Also can i use MessageFormat in any way to solve above problem.

Something like the below works:
String str = "We’ve added {gift} coupon worth {price}. Enjoy!";
System.out.println(str);
str = str.replace("{gift}", "unicorns");
str = str.replace("{price}", "$399");
System.out.println(str);
You can't use a MessageFormat to replace text in the input String you've provided, but if you save your String as a format to use it MessageFormat, it makes it a lot easier to read:
String defaultFormat = "We’ve added {0} coupon worth {1}. Enjoy!";
String defaultOutput = MessageFormat.format(defaultFormat, "unicorns", "$399");
System.out.println(defaultOutput);
defaultOutput = MessageFormat.format(defaultFormat, "leprechaun", "$199");
System.out.println(defaultOutput);
String yodaOutput = MessageFormat.format("coupon worth {1} we have added for {0}. Enjoy!", "unicorns", "$399");
System.out.println(yodaOutput);